solve 2cos^2x = square root of three cosx (couldnt find the sign on keyboard sorry) for principal values of x. express the solutions in degrees
2 cos^2 x = sqrt (3 cos x)
Note: cos x must be + to have a real square root, so only in quadrants 1 and 4)
4 cos^4 x = 3 cos x
cos^3 x = 3/4
cos x = .75^(1/3)
cos x = .90856
x = 25.1 deg or 360-25.1 deg
Thank you very much!
To solve the equation 2cos^2(x) = √3cos(x), we can rearrange the terms to obtain a quadratic equation in terms of cos(x). Here's how:
1. First, let's move all terms to the left side of the equation:
2cos^2(x) - √3cos(x) = 0
2. Next, factor out the common term cos(x):
cos(x)(2cos(x) - √3) = 0
3. Now, we have two possible solutions:
a) cos(x) = 0
b) 2cos(x) - √3 = 0
Let's solve each of these separately:
a) For cos(x) = 0:
To find the principal values of x, we need to determine when the cosine function equals zero. This occurs at x = π/2 + nπ and x = 3π/2 + nπ, where n is an integer. However, we are looking for the solutions in degrees, so we convert the radian solutions to degrees:
x = 90° + n(180°) and x = 270° + n(180°).
b) For 2cos(x) - √3 = 0:
To solve this equation for cos(x), isolate the cosine term:
2cos(x) = √3
cos(x) = √3/2
To find the principal values of x for cos(x) = √3/2, we can refer to the unit circle. At particular angles, the cosine function equals √3/2. The principal values occur at x = 30° + n(360°) and x = 330° + n(360°), where n is an integer.
Hence, the principal values of x for the given equation 2cos^2(x) = √3cos(x) are:
x = 90° + n(180°)
x = 270° + n(180°)
x = 30° + n(360°)
x = 330° + n(360°)
where n represents any integer.